Let i be the angle of incidence at the interface of Region III and Region IV. Using shell’s law for Region I and Region III, we get

n0sinθ =

sini For a total internal reflection in Region III, we have

i ≥

iC (Sin i) ≥

() (sin i) ≥

() Therefore,

n0sinθ =

× =

⇒ sin θ =

⇒ θ =

sin−1()